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The bilateral shift $(Tx{)}_n=x_{n+1}$ is unitary. It has empty point spectrum, while its approximate point spectrum is the unit circle. For $|\lambda |=1$, normalized vectors with entries ${\lambda }^n$ on an interval of length $N$ fail to satisfy $Tx=\lambda x$ only at the two endpoints, giving residual norm $\sqrt{2/N}$.

It is the two-sided sequence case of a shift operator.

The spectrum of a bounded operator consists of $\lambda$ for which $T-\lambda I$ has no bounded two-sided inverse. The point spectrum consists of eigenvalues, so $\ker (T-\lambda I)\ne 0$. The approximate point spectrum consists of $\lambda$ admitting unit vectors $x_n$ with $\Vert (T-\lambda I)x_n\Vert \to 0$; equivalently $T-\lambda I$ is not bounded below.

If $T-\lambda I$ is invertible, its inverse bounds it below and its Banach-space adjoint is invertible. Thus neither alternative on the right of the claimed union occurs. Conversely suppose $\lambda$ belongs to neither term and write $S=T-\lambda I$. There is $c>0$ with $\Vert Sx\Vert \ge c\Vert x\Vert$, so $S$ is injective and has closed range: convergence of $S{x}_n$ makes $x_n$ Cauchy, and completeness supplies a preimage of its limit. Also $\ker S^{\prime }=0$. The allowed annihilator identity makes the range dense, hence equal to $X$. Its inverse has norm at most $1/c$. This proves the spectrum decomposition through the Banach-space adjoint:For the right shift on ${\ell }^1$, $\Vert T\Vert =1$, so the allowed norm bound puts its spectrum in the closed unit disc. Under the linear pairing with ${\ell }^{\infty }$, its Banach-space adjoint is the left shift $T^{\prime }(b_1,b_2,\dots )=(b_2,b_3,\dots )$. For each $|\lambda |\le 1$ the bounded nonzero sequence $(1,\lambda ,{\lambda }^2,\dots )$ is an eigenvector of $T^{\prime }$ with eigenvalue $\lambda$. The union identity gives the reverse inclusion. Hence $\sigma (T)=\{\lambda :|\lambda |\le 1\}$. This Banach-space adjoint uses the complex-linear dual, so its eigenvalue is $\lambda$ itself.

The spectrum of a bounded operator is $\sigma (T)=\{\lambda \in \mathbb{C}:T-\lambda I\text{ has no bounded everywhere-defined inverse}\}$. Initially define $T$ on finite linear combinations of the orthonormal basis. The bilateral shift operator $S{e}_n=e_{n+1}$ preserves the Hilbert space norm and extends to a unitary operator on $H$; its inverse is $S^{\ast }{e}_n=e_{n-1}$. Thus $T=S+S^{\ast }$ extends uniquely by density, is bounded with norm at most two, and is self-adjoint.

For $v_N=(2N+1{)}^{-1/2}{\sum }_{n=-N}^N{e}_n$, $\Vert v_N\Vert =1$ and the only nonzero coefficients of $(T-2I)v_N$ occur at $-N-1,-N,N,N+1$. HenceIt follows that $\Vert T{v}_N\Vert \to 2$, proving $\boxed{\Vert T\Vert =2}$. Also $\boxed{2\in \sigma (T)}$: a bounded inverse of $T-2I$ would give $1\le \Vert (T-2I{)}^{-1}\Vert \Vert (T-2I)v_N\Vert \to 0$. Thus two belongs to the approximate point spectrum.

An eigenvector $x=\sum x_n{e}_n$ would satisfy $\lambda x_n=x_{n-1}+x_{n+1}$. Self-adjointness and the norm bound force $\lambda \in [-2,2]$. If $|\lambda |<2$, write $\lambda =2\cos \theta$ with $0<\theta <\pi$; the general complex solution of the linear recurrence relation isThe Cesàro mean of $|x_n{|}^2$ over $n=-N,\dots ,N$ tends to $|A{|}^2+|B{|}^2$, since the mixed geometric sum divided by $2N+1$ tends to zero. For an ${\ell }^2$ sequence the same mean tends to zero. Therefore $A=B=0$. At $\lambda =\pm 2$, the general solution is $(A+Bn)(\pm 1{)}^n$, again square summable only when zero. This also covers $\lambda =0$ through $\theta =\pi /2$. Thus $\boxed{T\text{ has no eigenvectors}}$.

Finally the bounded sequence $e_{3j}$ has images $T{e}_{3j}=e_{3j-1}+e_{3j+1}$ with pairwise disjoint supports. Distinct images have distance two, so no subsequence converges. Hence $\boxed{T\text{ is not compact}}$.

For $x\in X$,The map $S$ defined by $(Sx{)}_n=x_{n-1}$ satisfies $ST=TS=I$, so $T$ is a surjective isometry and hence a unitary operator.

If $Tx=\lambda x$, thenFor $\lambda \ne 0$, this gives $x_n={\lambda }^n{x}_0$ for every integer $n$. When $|\lambda |>1$ the positive tail is not square summable, when $|\lambda |<1$ the negative tail is not square summable, and when $|\lambda |=1$ neither tail is square summable unless $x_0=0$. The case $\lambda =0$ also forces every coordinate to vanish. Thus

The reverse triangle inequality gives, for every $x$,Hence $T-\lambda I$ is bounded below whenever $|\lambda |\ne 1$, so such $\lambda$ do not belong to the approximate point spectrum.

Now let $|\lambda |=1$ and define the unit vectorThe equation $x_{n+1}^{(N)}=\lambda x_n^{(N)}$ fails only at the two endpoints, soTherefore the bilateral shift operator has

For a bounded linear operator on a complex Banach space, either a spectral scalar has approximate unit eigenvectors or it is an eigenvalue of the Banach-space adjoint. Indeed outside the approximate point spectrum, $S=T-\lambda I$ is bounded below, hence injective with closed range. If the adjoint has no zero vector in its kernel other than zero, the Hahn-Banach theorem makes the range dense: any proper closed range would have a nonzero continuous functional annihilating it. Thus the range is all of the space and the lower bound gives a bounded inverse. Conversely invertibility rules out both alternatives. With complex-linear duals there is no conjugation of $\lambda$ in the displayed formula; Hilbert-adjoint conventions must be distinguished.